Solve this

Given Equation :

$(n+3) !=56 \times(n+1) !$

To Find : Value of $n$

Formula: $n !=n \times(n-1) !$

By given equation,

$(n+3) !=56 \times(n+1) !$

By using above formula we can write,

$\therefore(n+3) \times(n+2) \times(n+1) !=56 \times(n+1) !$

Cancelling the term $(n+1) !$ from both the sides,

$\therefore(n+3) \times(n+2)=56$

$\therefore(n+3) \times(n+2)=(8) \times(7)$

Comparing both the sides, we get

$\therefore \mathrm{n}=5$

Conclusion : Value of $n$ is $5 .$

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 56 into product of two consecutive numbers and then by observing two sides of equation we can get value of $n$.